151028 总结
T1
貌似很水，求一个最长的序列使得阶乘和等于当前序列，我们发现对于每个数都有独立性，那么把1~9特判掉就好啦 100
T2
看到数据范围大致明白是搜索了，我觉得宽搜状态不好设置所以写的深搜
剪枝：
1.层数限制，大于100层（大致）剪掉
2.不能让人在原地绕圈
3.把箱子不可能达到的地点预处理出来，这样可以O(1)判断
4.只有3个箱子、、等等只有3个箱子？我们完全可以把箱子分开设置状态啊
一共7*7,除去边界只有5个格子，4个物体所以5^7状态，大概也就80000种状态，我们记忆化一下然后再搞？
然后过了样例，但是大数据有时候会卡比如
1111111
1000001
1004001
1320001
1320001
1320001
1111111
然后还要加启发式优化，如果当前状态严格不优大胆剪掉就好啦
最后100
T3
其实这道题比T2简单的= =
二分边权，把所有白边加上当前边权，每次kruskal和k比较就好啦
因为脑残想到了LCT所以写了暴力，然后暴力严重挂掉导致只有10分
贴上代码
//Copyright(c)2015 liuchenrui
#include<cstdio>
#include<ctime>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<map>
using namespace std;
int ans[1001],t;
int n;
char s[1001];
int main()
{
freopen("function.in","r",stdin);
freopen("function.out","w",stdout);
cin>>n;scanf("%s",s+1);
for(int i=1;i<=n;i++)
{
if(s[i]=='0')continue;
if(s[i]=='1')continue;
if(s[i]=='2')ans[++t]=2;
if(s[i]=='3')ans[++t]=3;
if(s[i]=='4')ans[++t]=2,ans[++t]=2,ans[++t]=3;
if(s[i]=='5')ans[++t]=5;
if(s[i]=='6')ans[++t]=3,ans[++t]=5;
if(s[i]=='7')ans[++t]=7;
if(s[i]=='8')ans[++t]=2,ans[++t]=2,ans[++t]=2,ans[++t]=7;
if(s[i]=='9')ans[++t]=2,ans[++t]=3,ans[++t]=3,ans[++t]=7;
}
sort(ans+1,ans+t+1);
for(int i=t;i>=1;i--)printf("%d",ans[i]);
}
//Copyright(c)2015 liuchenrui
#include<cstdio>
#include<ctime>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<conio.h>
#include<map>
#define inf 1000000000
#define rep() for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)
inline void R(int &v)
{
v=0;char c=0;int p=1;
while(c<'0' || c>'9'){if(c=='-')p=-1;c=getchar();}
while(c>='0' && c<='9'){v=(v<<3)+(v<<1)+c-'0';c=getchar();}
v*=p;
}
char map[11][11];
bool can[11][11];
char g[11][11];
int n,m;
int stax,stay;
int num;
const int fx[]={0,1,0,-1,0};
const int fy[]={0,0,1,0,-1};
int ans=inf;
int zt[500000];
const int zy[]={1,5,25,125,625,3125,15625,78125};
void huifu(int x,int y)
{
if(g[x][y]=='3')map[x][y]='3';
else map[x][y]='0';
}
void print(int step)
{
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
{
putchar(map[i][j]);
}
putchar(10);
}
printf("step=%d\n",step);
putchar(10);
}
int calc()
{
int t=1,ret=0;bool flag=true;
rep()
{
if(map[i][j]=='4')ret+=zy[0]*(i-2)+zy[1]*(j-2);
if(map[i][j]=='2')ret+=zy[t<<1]*(i-2)+zy[t<<1|1]*(j-2),t++;
if(map[i][j]=='2' && g[i][j]!='3')flag=false;
}
if(flag)return -1;
return ret;
}
void dfs(int nowx,int nowy,int step)
{
int hahaha=calc();
if(hahaha>0)if(step>=zt[hahaha])return;
if(hahaha>0)zt[hahaha]=step;
//print(step);
//getch();
if(step>100)return;
if(hahaha==-1)
{
ans=std::min(ans,step);
return;
}
for(int i=1;i<=4;i++)
{
int px=nowx+fx[i];
int py=nowy+fy[i];
if(map[px][py]=='1')continue;
if(map[px][py]=='0' || map[px][py]=='3')
{
map[px][py]='4';
huifu(nowx,nowy);
dfs(px,py,step+1);
huifu(px,py);
map[nowx][nowy]='4';
}
else if(map[px][py]=='2')
{
int rx=px+fx[i];
int ry=py+fy[i];
if(!can[rx][ry])continue;
if(map[rx][ry]=='2')continue;
map[rx][ry]='2';
map[px][py]='4';
huifu(nowx,nowy);
dfs(px,py,step+1);
map[nowx][nowy]='4';
map[px][py]='2';
huifu(rx,ry);
}
}
}
int main()
{
freopen("box.in","r",stdin);
freopen("box.out","w",stdout);
memset(can,true,sizeof(can));
memset(zt,63,sizeof(zt));
R(n),R(m);
for(int i=1;i<=n;i++)
{
scanf("%s",map[i]+1);
}
rep()g[i][j]=map[i][j];
rep()
{
if(map[i][j]=='1')can[i][j]=false;
if(map[i+1][j]=='1' && map[i][j+1]=='1')can[i][j]=false;
if(map[i-1][j]=='1' && map[i][j+1]=='1')can[i][j]=false;
if(map[i+1][j]=='1' && map[i][j-1]=='1')can[i][j]=false;
if(map[i-1][j]=='1' && map[i][j-1]=='1')can[i][j]=false;
if(map[i][j]=='3')can[i][j]=true,num++;
if(map[i][j]=='4')stax=i,stay=j;
}
dfs(stax,stay,0);
std::cout<<ans<<std::endl;
}
//Copyright(c)2015 liuchenrui
#include<cstdio>
#include<ctime>
#include<iostream>
#include<algorithm>
#include<cstring>
#define inf 1000000000
using namespace std;
inline void R(int &v)
{
v=0;char c=0;int p=1;
while(c<'0' || c>'9'){if(c=='-')p=-1;c=getchar();}
while(c>='0' && c<='9'){v=(v<<3)+(v<<1)+c-'0';c=getchar();}
v*=p;
}
int father[100010];
int sum;
int from[100010],to[100010],len[100010],color[100010];
struct Edge
{
int from,to,len,color;
}edge[200010];
int getfather(int v)
{
if(father[v]==v)return v;
father[v]=getfather(father[v]);
return father[v];
}
int n,m,k;
int size;
void addedge(int x,int y,int z,int color)
{
size++;
edge[size].from=x;
edge[size].to=y;
edge[size].len=z;
edge[size].color=color;
}
bool comp1(const Edge &a,const Edge &b)
{
if(a.len!=b.len)return a.len<b.len;
return a.color<b.color;
}
bool comp2(const Edge &a,const Edge &b)
{
if(a.len!=b.len)return a.len<b.len;
return a.color>b.color;
}
bool kruskal(int num)
{
size=0;
for(int i=0;i<=n;i++)father[i]=i;
for(int i=1;i<=m;i++)
{
if(color[i]==0)addedge(from[i],to[i],len[i]+num,0);
else addedge(from[i],to[i],len[i],1);
//cerr<<from[i]<<" "<<to[i]<<endl;
}
sort(edge+1,edge+size+1,comp1);
int ret1=0;sum=0;
for(int i=1;i<=size;i++)
{
int l=getfather(edge[i].from);
int r=getfather(edge[i].to);
if(l!=r)
{
if(!edge[i].color)ret1++;
father[l]=r;
if(edge[i].color==1 || ret1>k)sum+=edge[i].len;
else sum+=edge[i].len-num;
}
}
return ret1>=k;
}
int main()
{
freopen("tree.in","r",stdin);
freopen("tree.out","w",stdout);
R(n),R(m),R(k);
for(int i=1;i<=m;i++)
{
R(from[i]),R(to[i]),R(len[i]),R(color[i]);
from[i]++,to[i]++;
}
n++;n++;
int l=-101,r=101;
//cerr<<kruskal(-4)<<endl;
//cerr<<sum<<endl;
//return 0;
while(l<r)
{
int mid=l+r>>1;
if(kruskal(mid))l=mid+1;
else r=mid;
}
/*for(int i=-100;i<=100;i++)
{
int u=kruskal(i);
fprintf(stderr,"%d %d %d\n",i,u,sum);
}*/
kruskal(l-1);
std::cout<<sum<<std::endl;
}